2.1.13 Problem 13

Internal problem ID [9183]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 13
Date solved : Friday, April 25, 2025 at 05:59:16 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{t}^2{y}^{\prime \prime }-t(t+2)y^{\prime }+(t+2)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.112 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& t^2{y}^{\prime \prime }+(-t^2-2t)y^{\prime }+(t+2)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =t^2\\ \text{(3)}& B& =-t^2-2tC& =t+2\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(t)& =y{e}^{\int \frac{B}{2A}dt}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(t)=rz(t)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{1}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =1\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\frac{z\left(t\right)}{4}\end{array}$$

Equation (7) is now solved. After finding $z(t)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(t\right)e^{-\int \frac{B}{2A}dt}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.13: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0-0\\ & =0\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=\frac{1}{4}$ is not a function of $t$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(t)={\mathrm{e}}^{-\frac{t}{2}}$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-t^2-2t}{t^2}dt}\\ & =z_1{e}^{\frac{t}{2}+\ln \left(t\right)}\\ & =z_1\left(t{\mathrm{e}}^{\frac{t}{2}}\right)\end{array}$$

Which simplifies to

$$y_1=t$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dt}}{y_1^2}dt$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-t^2-2t}{t^2}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{t+2\ln \left(t\right)}}{{\left(y_1\right)}^2}dt\\ & =y_1\left(\frac{{\mathrm{e}}^{t+2\ln \left(t\right)}}{t^2}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(t\right)+c_2\left(t\left(\frac{{\mathrm{e}}^{t+2\ln \left(t\right)}}{t^2}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.004 (sec). Leaf size: 12

ode:=t^2*diff(diff(y(t),t),t)-t*(t+2)*diff(y(t),t)+(t+2)*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 

$$y=t({\mathrm{e}}^t{c}_2+c_1)$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & t^2\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)-t(t+2)\left(\frac{d}{dt}y\left(t\right)\right)+(t+2)y\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)=-\frac{(t+2)y\left(t\right)}{t^2}+\frac{(t+2)\left(\frac{d}{dt}y\left(t\right)\right)}{t}\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)-\frac{(t+2)\left(\frac{d}{dt}y\left(t\right)\right)}{t}+\frac{(t+2)y\left(t\right)}{t^2}=0\\ ◻& & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(t\right)=-\frac{t+2}{t},P_3\left(t\right)=\frac{t+2}{t^2}]\\ & \circ & t\cdot P_2\left(t\right)\text{is analytic at}t=0\\ & & (t\cdot P_2\left(t\right))|{}_{t=0}=-2\\ & \circ & t^2\cdot P_3\left(t\right)\text{is analytic at}t=0\\ & & (t^2\cdot P_3\left(t\right))|{}_{t=0}=2\\ & \circ & t=0\text{is a regular singular point}\\ & & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & & t_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & t^2\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)-t(t+2)\left(\frac{d}{dt}y\left(t\right)\right)+(t+2)y\left(t\right)=0\\ \bullet & & \text{Assume series solution for}y\left(t\right)\\ & & y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{t}^m\cdot y\left(t\right)\text{to series expansion for}m=0..1\\ & & t^m\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & t^m\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{t}^{k+r}\\ & \circ & \text{Convert}{t}^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)\text{to series expansion for}m=1..2\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)t^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)t^{k+r}\\ & \circ & \text{Convert}{t}^2\cdot \left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)\text{to series expansion}\\ & & t^2\cdot \left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)t^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0(-1+r)(-2+r)t^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_k(k+r-1)(k+r-2)-a_{k-1}(k+r-2))t^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & (-1+r)(-2+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{1,2\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k+r-2)(a_k(k+r-1)-a_{k-1})=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & (k+r-1)(a_{k+1}(k+r)-a_k)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k}{k+r}\\ \bullet & & \text{Recursion relation for}r=1\\ & & a_{k+1}=\frac{a_k}{k+1}\\ \bullet & & \text{Solution for}r=1\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+1},a_{k+1}=\frac{a_k}{k+1}]\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+1}=\frac{a_k}{k+2}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+2},a_{k+1}=\frac{a_k}{k+2}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(t\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+1})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{t}^{k+2}),a_{k+1}=\frac{a_k}{k+1},b_{k+1}=\frac{b_k}{k+2}]\end{array}$$

✓ Mathematica. Time used: 0.041 (sec). Leaf size: 17

ode=t^2*D[y[t],{t,2}]-t*(t+2)*D[y[t],t]+(t+2)*y[t] == 0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to et(c_2{e}^t+c_1)$$

✓ Sympy. Time used: 0.857 (sec). Leaf size: 29

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t**2*Derivative(y(t), (t, 2)) - t*(t + 2)*Derivative(y(t), t) + (t + 2)*y(t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=C_2{t}^2(\frac{t^3}{24}+\frac{t^2}{6}+\frac{t}{2}+1)+C_{1}t+O\left(t^6\right)$$